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If we have only 3 colors for 5 boards, there are only 2 possible options 3 (first color) + 1 (second color) + 1 (third color) and 2(first color) + 2 (second color) + 1 (third color). There is 3 different ways for each option.

For example, option 1 if we have red, green, blue:

red + green + blue + blue + blue
red + green + green + green + blue
red + red + red + green + blue

option 2:

red + red + green + green + blue
red + red + green + blue + blue
red + green + green + blue + blue

So, \(C^3_1\) actually means that we choose one color out of 3 for 3 boards (option 1) or for 1 board (option 2)
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For 3 colors: one color of 3 for 3+1+1 combination and one color of 3 for 2+2+1 combination.

Can we write for the option 2 as 6C3 * 3C2 (which is equal to 6C3 * 3C1) ?

This is because let us say we have chosen 3 colors for 3 boards, and now we need to choose 2 colors out of 3 for the remaining 2 boards.

So option is 6C3 * 3C1 + 6C3 * 3C2.
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Yup, \(C^3_2\) is correct too.
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good question indeed.
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10 seconds approach:
Essentially the question says, there are 5 balls and 6 kids - blue, green, beige, white, yellow, and red.
How can we distribute these 5 balls among these 6 kids.
Mathematically, this will look like
a+b+c+d+e+f = 5
where each variable represent a color and 5 represents the package of poster boards.
Now your task is to divide the 5 balls into 6 different groups.
For example, lets say there are 2 kids and I want to split 5 balls among them.
For this purpose, I will use a stick/partition to divide these 5 balls into 2 groups. Visualise as below:
case-1: OO | OOO
case-2: O | OOOO
Case-3: | OOOO
Case-4: OOOO|O
Case-5: OOO | OOO
Case-6: OOOOO |
So essentially, I am arranging 6 objects in a row of which 5 are identical.
which is equal to 6!/5! = 6

So now, our main question reduces to arranging 10 objects of which 5 each are identical.
Answer = 10!/ 5!5! = 252

This is a question on partition rule in PnC. Once you identify the concept applicable, this should take less than 10 seconds to solve.
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